传送门——BZOJCH
传送门——Vjudge
设\(f_i\)表示\(i\)到\(1\)号点的最短距离，\(g_i\)表示\(i\)到\(2\)号点的最短距离，\(s_i\)表示\(n+1\)号点到\(i\)号点的最短距离，\(A=s_1,B=s_2\)
根据最短路三角形不等式，\(|f_i - A| \leq s_i \leq f_i + A , |g_i - B| \leq s_i \leq g_i + B\)
而\(s_i\)要取到最小值，所以\(s_i = \max\{|f_i - A| , |g_i - B|\}\)
所以我们要求的是\(\sum\limits_{i=1}^N \max\{|f_i - A| , |g_i - B|\}\)，这相当于求一个动点\((A,B)\)到平面上\(N\)个点\((f_i,g_i)\)的最小切比雪夫距离和。
切比雪夫距离可以转为曼哈顿距离，将坐标\((x,y)\)变为\((\frac{x+y}{2} , \frac{x-y}{2})\)，前者的切比雪夫距离等效于后者的曼哈顿距离。而曼哈顿距离可以直接拆开横纵坐标然后取中位数。
注意：我天真的以为2012年的题不会卡SPFA……
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<queue>
#define INF 0x3f3f3f3f
//This code is written by Itst
using namespace std;inline int read(){int a = 0;char c = getchar();while(!isdigit(c) && c != EOF)c = getchar();while(isdigit(c)){a = a * 10 + c - 48;c = getchar();}return a;
}#define PLI pair < long long , int >
#define st first
#define nd second
const int MAXN = 1e5 + 7;
struct Edge{int end , upEd , w;
}Ed[MAXN * 6];
int head[MAXN] , N , M , cntEd;
long long dis[2][MAXN];
priority_queue < PLI > q;inline void addEd(int a , int b , int w){Ed[++cntEd].end = b;Ed[cntEd].w = w;Ed[cntEd].upEd = head[a];head[a] = cntEd;
}void SPFA(int ind){memset(dis[ind] , 0x3f , sizeof(long long) * (N + 1));dis[ind][ind + 1] = 0;q.push(PLI(0 , ind + 1));while(!q.empty()){PLI t = q.top();q.pop();if(-t.st != dis[ind][t.nd]) continue;for(int i = head[t.nd] ; i ; i = Ed[i].upEd)if(dis[ind][Ed[i].end] > dis[ind][t.nd] + Ed[i].w){dis[ind][Ed[i].end] = dis[ind][t.nd] + Ed[i].w;q.push(PLI(-dis[ind][Ed[i].end] , Ed[i].end));}}
}inline long long abss(long long x){return x < 0 ? -x : x;}void out(long long a , int b){cout << a / b << '.';a %= b;for(int i = 1 ; i <= 8 ; ++i){a *= 10;cout << a / b;a %= b;}putchar('\n');
}int main(){vector < long long > x , y;for(int T = read() ; T ; --T){N = read(); M = read();memset(head , 0 , sizeof(int) * (N + 1));cntEd = 0;for(int i = 1 ; i <= M ; ++i){int a = read() , b = read() , c = read();addEd(a , b , c); addEd(b , a , c);}SPFA(0); SPFA(1);x.clear(); y.clear();long long sum = 0;for(int i = 1 ; i <= N ; ++i){x.push_back(dis[0][i] - dis[1][i]);y.push_back(dis[0][i] + dis[1][i]);}sort(x.begin() , x.end()); sort(y.begin() , y.end());long long mid = x[N >> 1];for(int i = 0 ; i < N ; ++i)sum += abss(x[i] - mid);mid = y[N >> 1];for(int i = 0 ; i < N ; ++i)sum += abss(y[i] - mid);out(sum , 2 * N);cerr << N << ' ' << sum << endl;}return 0;
}
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